A particle is movin along straight line such that dot product of its acceleration `veca` and velocity `vecv` is negative. How many times particle can be found at a distance d from the origin?

Options

A. One
B. Three
C. Two
D. Can't be determined with given data

Explanation

To solve the problem, we need to analyze the motion of the particle given that the dot product of its acceleration \(\vec{a}\) and velocity \(\vec{v}\) is negative. This implies that the angle \(\theta\) between the vectors \(\vec{a}\) and \(\vec{v}\) is greater than 90 degrees.

### Step-by-Step Solution:

1. **Understanding the Dot Product**:
The dot product of two vectors is given by:
\[ \vec{a} \cdot \vec{v} = |\vec{a}| |\vec{v}| \cos(\theta) \]
Since the dot product is negative, we have:
\[ \cos(\theta) < 0 \implies \theta > 90^\circ \]
This means that the acceleration is acting in the opposite direction to the velocity.

2. **Implication of Negative Dot Product**:
When the acceleration is opposite to the velocity, it means that the particle is decelerating. The particle will eventually come to a stop when its velocity becomes zero.

3. **Motion Analysis**:
Let's denote the initial position of the particle as \(x_0\) and the distance from the origin as \(d\). As the particle moves, it will decelerate until it stops. After reaching the maximum displacement, it will start moving back towards the origin.

4. **Determining the Path**:
The particle can be found at a distance \(d\) from the origin twice:
- Once while moving away from the origin (before it stops).
- Once while returning back towards the origin (after it has reversed direction).

5. **Conclusion**:
Since the particle can reach the distance \(d\) twice during its motion (once while moving away and once while returning), the answer to the question is:
\[ \text{The particle can be found at a distance } d \text{ from the origin } 2 \text{ times.} \]